\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\)

\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{440}\)

\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{20.22}\)

\(A=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{20}-\frac{1}{22}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{22}\right)=2.\frac{5}{11}=\frac{10}{11}\)

\(B=\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{99}{100}\)

\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{9.11}{10.10}\)

\(B=\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}\cdot\frac{3\cdot4\cdot5\cdot...\cdot11}{2\cdot3\cdot4\cdot....\cdot10}\)

\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)

c)x:25/8-3/4=9/4

x:25/8=9/4+3/4

x:25/8=3

x=3 nhân 25/8

x=75/8

tất cả các bài có người làm rồi li-ke cho mình nha

1/1.2+1/3.4+1/5.6+...+1/49.50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=1/1+1/2+1/3+1/4+...+1/49+1/50-2(1/2+1/4+1/6+...+1/50)

=1/1+1/2+1/3+1/4+...+1/49+1/50-(1/1+1/2+1/3+1/4+...+1/25)

=1/26+1/27+...+1/50=1/26+1/27+...+1/50(đpcm)

b. 1/1-1/2+1/3-1/4+...+1/99-1/100=99/100

7/12=175/300; 5/6=10/12=250/300; 99/100=297/300

(hình như khúc này đề bài sai hả bạn) bạn tự tính ra nhé

bài 2: a.x+1/10+x/12+x/14+...x+1/20

(x+x+x...+x)+(1/10+1/12+...+1/20)

ko có kết quả sao tìm x được bạn:[

b.x+1/2000+x+2/1999=x+3/1998+x+4/1997

x+1/2000+x+2/1999=x+3/1998+x+4/1997

(x+1/2000+1)+(x+2/1999+1)=(x+3/1998+1)+(x+4/1997+1)

x+2002/2000+x+2002/1999=x+2002/1998+x+2002/1997

x+2002(1/2000+1/1999)=(x+2002)(1/1998+1/1997)

=>(1/2000+1/1999)=(1/1998+1/1997)

x+2002(1/2000+1/1999)-(x+2002)(1/1998+1/1997)=0

(x+2002)(1/2000+1/1999-1/1998-1/1997)=0

(x+2002).0=0

(x+2002)=0

x =0-2002=-2002

Chúc bạn học tốt.

Chỗ 4 mũ 2/3.5 x ... x 59 mũ 2/58.60 nha

a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)

                                                                                   \(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)

=> đpcm

Study well ! >_<

c, Ta có: \(2.|\dfrac{1}{2}x - \dfrac{1}{3}| - \dfrac{3}{2} = \dfrac{1}{4}\)

\(\Rightarrow\) \(2.|\dfrac{1}{2}x - \dfrac{1}{3}| = \dfrac{7}{4}\)

\(\Rightarrow\) \(|\dfrac{1}{2}x - \dfrac{1}{3}| = \dfrac{7}{8}\)

\(\Rightarrow\) \(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{29}{24}\\\dfrac{1}{2}x=-\dfrac{13}{24}\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x=\dfrac{29}{24}\\x=-\dfrac{13}{12}\end{matrix}\right.\)